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Electronic Journal of Qualitative Theory of Differential Equations 2006, No. 7, 1-14;http://www.math.u-szeged.hu/ejqtde/

Positive Solutions of Second Order Boundary Value Problems With Changing Signs Carath´eodory

Nonlinearities

Abdelkader Boucherif

King Fahd University of Petroleum and Minerals Department of Mathematical Sciences, Box 5046

Dhahran, 31261, Saudi Arabia e-mail: aboucher@kfupm.edu.sa

Johnny Henderson Department of Mathematics

Baylor University, Waco, TX 76798-7328, USA e-mail: Johnny Henderson@baylor.edu

Abstract

In this paper we investigate the existence of positive solutions of two-point boundary value problems for nonlinear second order differential equations of the form (py0)0(t) +q(t)y(t) =f(t, y(t), y0(t)), wheref is a Carath´eodory function, which may change sign, with respect to its second argument, infinitely many times.

2000 Mathematics Subject Classification: 34B15, 34B18

Keywords: positive solutions, Carath´eodory function, topological transversality the- orem, upper and lower solutions.

1 Introduction

We are interested in the existence of positive solutions of the two-point boundary value problem,

(py0)0(t) +q(t)y(t) =f(t, y(t), y0(t)), 0< t <1,

y(0) =y(1) = 0. (1.1)

corresponding author

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Problems of this type arise naturally in the description of physical phenomena, where only positive solutions, that is, solutions y satisfying y(t)>0 for allt∈(0,1), are meaningful. It is well known that Krasnoselskii’s fixed point theorem in a cone has been instrumental in proving existence of positive solutions of problem (1.1). Most of the previous works deal with the case p(t) = 1, q(t) = 0, for allt ∈[0,1],and assume that f is nonnegative, that f does not depend on y0, and that f : [0,1]×[0,+∞)→ [0,+∞) is continuous and satisfies, either

lim inf

u→0+ min

0≤t≤1

f(t, u)

u = +∞, and lim sup

u→+∞

0≤t≤1max

f(t, u)

u = 0 (sublinear case), or

lim sup

u→o+ max

0≤t≤1

f(t, u)

u = 0, and lim inf

u→+∞ min

0≤t≤1

f(t, u)

u = +∞ (superlinear case).

See for instance [1], [7], [12], [13], [16] and the references therein. The above conditions have been relaxed in [17] and [18], where the authors remove the condition f is nonnegative, and they consider the behavior of f relative to π2. Notice that π2 is the first eigenvalue of the operator u → −u00, subject to the boundary conditions in (1.1). The arguments in [17] and [18] are based on fixed point index theory in a cone. When the nonlinear term depends also on the first derivative of y, we refer the interested reader to [2], [3], [19]. The authors in [2], [8] and [20] deal with a singular problem. Several papers are concerned with the problem of the existence of multiple solutions. See for instance [3], [13], [14], [15] and [21]. However, our assumptions are simple and more general. In fact, we obtain a multiplicity result as a byproduct of our main result, with no extra assumptions. We exploit the fact that the nonlinearity changes sign with respect to its second argument. We do not rely on cone preserving mappings. Also, the sign of the Green’s function of the corresponding linear homogeneous problem plays no role in our study. However, we assume the existence of positive lower and upper solutions. We provide an example to motivate our assumptions. Our results complement and generalize those obtained in [21].

2 Topological Transversality Theory

In this section, we recall the most important notions and results related to the topo- logical transversality theory due to Granas. See Granas-Dugundji [10] for the details of the theory.

Let X be a Banach space, C a convex subset of X and U an open set inC.

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(i) g :X →X is compact if g(X) is compact

(ii) H : [0,1]× X → X is a compact homotopy if H is a homotopy and for all λ∈[0,1],H(λ,·) : X→X is compact.

(iii) g :U → C is called admissible ifgis compact and has no fixed points on Γ =∂U. Let MΓ(U ,C) denote the class of all admissible maps fromU toC.

(iv) A compact homotopyH is admissible if, for eachλ∈[0,1],H(λ,·) is admissible.

(v) Two mappings g and h in MΓ(U ,C) are homotopic if there is an admissible homotopy H : [0,1]×U → C such thatH(0,·) =g and H(1,·) =h.

(vi) g ∈ MΓ(U ,C) is called inessential if there is a fixed point free compact map h:U → C such that g|Γ = h|Γ. Otherwise, g is called essential

Lemma 2.1 Let dbe an arbitrary point inU andg ∈ MΓ(U ,C)be the constant map g(x)≡d. Then g is essential.

Lemma 2.2 g ∈ MΓ(U ,C) is inessential if and only if g is homotopic to a fixed point free compact map.

Theorem 2.1 Let g, h ∈ MΓ(U ,C) be homotopic maps. Then g is essential if and only if h is essential.

3 Preliminaries

3.1 Function Spaces.

LetI denote the real interval [0,1], and let R+ denote the set of all nonnegative real numbers. For k = 0,1, . . . , Ck(I) denotes the space of all functions u : I → R, whose derivatives up to order k are continuous. For u ∈ Ck(I) we define its norm by kuk = Pk

i=0{max

u(i)(t)

: t ∈ I}. Equipped with this norm Ck(I) is a Banach space. When k = 0, we shall use the notation kuk0 for the norm of u ∈ C(I). Also C01(I) shall denote the space {y ∈ C1(I) : y(0) = y(1) = 0}. It can be easily shown that (C01(I), k·k) is a Banach space.

A real valued function f defined on I ×R2 is said to be an L1−Caratheodory function if it satisfies the following conditions

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(i) f(t,·) is continuous for almost all t∈I (ii) f(·, z) is measurable for allz ∈R2

(iii) for eachρ >0,there existshρ∈L1(I;R+) such thatkzkR2 ≤ρimplies|f(t, z)| ≤ hρ(t) for almost all t∈I.

3.2 A Linear Problem

Consider the following linear boundary value problem

(py0)0(t) +q(t)y(t) =h(t), t∈(0,1), y(0) = 0, y(1) = 0,

where the coefficient functionsp and q satisfy

(H0) p ∈C1(I), q ∈C(I), p(t) ≥ p0 >0 for all t ∈ I, q(t) ≤p0π2 with strict inequality on a subset of I with positive measure.

Lemma 3.1 Assume that (H0) is satisfied. Then for any nontrivial y ∈ C01(I), we have

Z 1

0

[p(t)y0(t)2−q(t)y(t)2]dt >0.

Proof. It follows from (H0) that Z 1

0

[p(t)y0(t)2−q(t)y(t)]dt > p0

Z 1

0

[y0(t)2−π2y(t)2]dt.

Consider the functional χ:C01(I)→R defined by χ(y) =

Z 1

0

[y0(t)2−π2y(t)2]dt.

Results from the classical calculus of variations (see [6]) shows that χ(y)≥ 0 for all y∈C01(I). Hence, the conclusion of the Lemma holds.

Lemma 3.2 If (H0) is satisfied, then the linear homogeneous problem (py0)0(t) + q(t)y(t) = 0, y(0) =y(1) = 0, has only the trivial solution.

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Proof. Assume on the contrary that the problem has a nontrivial solution y0.Then, we have

0 = Z 1

0

[(py00)0(t) +q(t)y0(t)]y0(t)dt.

But, Lemma 3.2 implies that Z 1

0

[(py00)0(t) +q(t)y0(t)]y0(t)dt=−[

Z 1

0

[(p(t)y00(t)2−q(t)y0(t)2]dt <0.

This contradiction shows thaty0 = 0,and the proof is complete.

As a consequence of Lemma 3.2, the corresponding Green’s function,G(t, s), exists and the linear nonhomogeneous problem (py0)0(t)+q(t)y(t) =h(t), y(0) =y(1) = 0, has a unique solution, given by y(t) =R1

0 G(t, s)h(s)ds.

Define a linear operator L:W2,1(I)∩C01(I)→L1(I) by Ly(t) := (py0)0(t) +q(t)y(t), t∈I.

It follows from the above thatLis one-to-one and onto, andL−1 :L1(I)→W2,1(I)∩ C01(I) is defined by

L−1h(t) :=

Z 1

0

G(t, s)h(s)ds.

By the compactness of the embedding W2,1(I) ,→ C1(I), the operator L−1 maps L1(I) into C01(I) and is compact.

4 Main Results

Consider the nonlinear problem (1.1)

(py0)0(t) +q(t)y(t) =f(t, y(t), y0(t)), t ∈(0,1), y(0) = 0, y(1) = 0.

The nonlinearity f : [0,1]×R2 →R is an L1-Carath´eodory function and satisfies (H1) There exist positive functions α ≤ β in C01(I) such that for allt ∈I, (i) Lα(t)≥f(t, α(t), α0(t)), Lβ(t)≤ f(t, β(t), β0(t))

(ii) f(t, α(t), α0(t))>0> f(t, β(t), β0(t))

(H2) there exist C > 0, Q ∈ L1(I;R+) and Ψ : [0,+∞) → [1,+∞) continuous and nondecreasing with 1

Ψ integrable over bounded intervals and R+∞

0

du

Ψ(u) = +∞,

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such that |f(t, y, z)| ≤Ψ (|z|) (Q(t) +C|z|) for all t∈[0,1], α≤y≤β, z ∈R. Remarks.

(i) α≤y≤β meansα(t)≤y(t)≤β(t) for allt ∈I.

(ii) Condition (H2) is known as a Nagumo-Wintner condition, and is more general than the usual Nagumo or Nagumo-Bernstein conditions.

Theorem 4.1 Assume (H1) and (H2) are satisfied. Then (1.1) has at least one positive solution y∈[α, β].

Proof. The proof will be given in several steps. Consider δ(y) = δ(α, y, β) = max(α,min(y, β)).

Since our arguments are based on the topological transversality theory, we consider the following one-parameter family of problems,

(py0)0(t) +q(t)y(t) =λf1(t, y(t), y0(t)), t∈(0,1),

y(0) = 0, y(1) = 0, (1.2λ)

where 0≤λ ≤1 and

f1(t, y(t), y0(t)) = f(t, δ y(t)), δ(y)0(t)

. (1.3)

Notice that (1.20) has only the trivial solution. Hence, we shall consider only the case 0< λ≤1.

Step.1. All possible solutions of (1.2λ) satisfyy(t)≤β(t) for all t∈I.

Let y 6= 0 be a possible solution of (1.2λ), and suppose on the contrary that there is a τ ∈ (0,1) such that y(τ) > β(τ). Then, there exists a maximal interval I1 = (a, b) such that τ ∈I1 and y(t)> β(t) for all t ∈I1.

It follows that δ(y(t)) =δ(α(t), y(t), β(t)) = β(t) and δ(y)0(t) = β0(t) for all t∈(a, b).

Let z(t) :=y(t)−β(t). Then, we have z(t)>0 for allt ∈I1. We have two possibilities.

(i) I1 ⊂ I. Then z(a) = 0, z(b) = 0 and z(t) >0 for all t ∈ (a, b). Hence, by (H1), for all t∈(a, b)

Lz(t) = Ly(t)−Lβ(t) = λf1(t, y(t), y0(t))−Lβ(t) =λf(t, δ y(t)), δ(y)0(t)

−Lβ(t)

≥λf(t, δ y(t)), δ(y)0(t)

− f(t, β(t), β0(t))

= (λ−1)f(t, β(t), β0(t))≥0,

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so that

Z

I1

Lz(t)z(t)dt= Z b

a

Lz(t)z(t)dt >0.

On the other hand, Lemma 3.2 implies that Z b

a

Lz(t)z(t)dt <0.

This is a contradiction.

(ii) I1 = I. In this case z(0) = 0, z(1) = 0 and z(t) > 0 for all t ∈ (0,1). It follows that

Z 1

0

Lz(t)z(t)dt >0, But again

Z 1

0

Lz(t)z(t)dt <0, and again, we arrive at a contradiction.

Therefore, we conclude

y(t)≤β(t) for allt ∈I.

Similarly, we can prove thatα(t)≤y(t) for all t∈I.

Hence, we have shown that any solution y of (1.2λ) satisfies

α(t)≤y(t)≤β(t) for allt ∈I. (1.4) But, for all y satisfying (1.4) f1 and f coincide. So, for λ = 1, we have that all solutions of (1.21) are solutions of (1.1). Moreover solutions of (1.2λ) satisfy the a priori bound, independently ofλ,

kyk0 ≤K0 :=kβk0.

Step.2. A priori bound on the derivative y0 for solutionsyof (1.2λ) satisfying the inequality (1.4).

Define K1 >0 by the formula RK1 0

du Ψ(u) > 1

p0[kQ0kL1 +2(C+kp0k0)K0], (this is possible because of the property of Ψ), where Q0(t) =Q(t) +kqk0K0.

We want to show that |y0(t)| ≤ K1 for all t ∈ I. Suppose, on the contrary that there exists τ1 such that |y01)| > K1. Then, there exists an interval [µ, ξ] ⊂ [0,1]

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such that the following situations occur:

(i) y0(µ) = 0, y0(ξ) =K1, 0< y0(t)< K1 µ < t < ξ, (ii) y0(µ) =K1, y0(ξ) = 0, 0< y0(t)< K1 µ < t < ξ, (iii) y0(µ) = 0, y0(ξ) =−K1, −K1 < y0(t)<0 µ < t < ξ,

(iv)y0(µ) =−K1, y0(ξ) = 0, −K1 < y0(t)<0 µ < t < ξ.

We study the first case. The others can be handled in a similar way. For (i), the differential equation in (1.2λ) and condition (H2) imply

|Ly(t)| ≤Ψ (y0(t)) (Q(t) +Cy0(t)) µ≤t≤ξ. (1.5)

Since

|Ly(t)|=|p(t)y00(t) +p0(t)y0(t) +q(t)y(t)|

≥ |p(t)y00(t)| − |p0(t)y0(t)| − |q(t)y(t)|, we have

|Ly(t)| ≥ |p(t)y00(t)| − kp0k0|y0(t)| − kqk0K0

≥p0y00(t)− kp0k0|y0(t)| − kqk0K0. It follows from (1.5) that

p0y00(t)≤Ψ (y0(t)) (Q(t) +Cy0(t)) +kp0k0|y0(t)|+kqk0K0

≤Q(t)Ψ (y0(t)) +Cy0(t)Ψ (y0(t)) +kp0k0|y0(t)|+kqk0K0. Notice that Ψ(z)≥1 for all z ≥0,so that

p0y00(t)≤Ψ (y0(t)) [(Q(t) +kqk0K0) +y0(t) (C+kp0k0)]

≤Ψ (y0(t)) (Q0(t) + (C+kp0k0)y0(t)). This implies

y00(t)

Ψ (y0(t)) ≤ 1

p0(Q0(t) + (C+kp0k0)y0(t)) for µ≤t≤ξ.

Integration from µ toξ, and a change of variables (see [8, Lemma A.10]) lead to

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Z K1

0

du Ψ(u) ≤ 1

p0

[kQ0kL1 + 2 (C+kp0k0)K0. This clearly contradicts the definition of K1.

Taking into consideration all the four cases above, we see that

|y0(t)| ≤K1 for all t∈I.

Let

K2 := max{K1,kα0k0,kβ0k0}.

Then, any solution y of (1.2λ) satisfying the inequality (1.4), is such that its first derivative y0 will satisfy the a priori bound

|y0(t)| ≤K2 for all t∈I.

As a consequence of Step 1 and Step 2 above, we deduce that any solution y of (1.2λ) satisfies

kyk ≤K :=K0+K2. (1.6) Since f is anL1-Caratheodory function, it follows from (1.6) that there exists hK ∈ L1(I : R+) such that |f(t, y(t), y0(t))| ≤ hK(t), for almost every t ∈ I. Now, the differential equation in (1.1) implies there exists φ ∈ L1(I : R+), depending on only p,||p0||0,||q||0, hK such that y00(t) ≤ φ(t) for almost every t ∈ I. In particular, y00 ∈L1(I :R+).

Step.3. Existence of solutions of (1.2λ).

IfG(t, s) is the Green’s function corresponding to the linear homogeneous problem (py0)0(t) +q(t)y(t) = 0, y(0) = 0 =y(1),then problem (1.2λ) is equivalent to

y(t) =λ Z 1

0

G(t, s)f1(s, y(s), y0(s))ds (1.7) Let

F1 :C1(I)→L1(I) be defined by, for all t ∈I,

F1(y)(t) =f1(t, y(t), y0(t)).

For all λ∈[0,1], t∈I, define

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H : [0,1]×C01(I)→C01(I) by

H(λ, y) =λL−1F1(y), i.e. H(λ, y)(t) =λ

Z 1

0

G(t, s)f1(s, y(s), y0(s))ds. (1.8) SinceL−1is compact andF1is continuous it follows thatH(λ,·) is a compact operator.

It is easily seen that H(·,·) is uniformly continuous inλ.

Let

U :={y∈C01(I) : kyk<1 +K}.

It is clear from Steps 1 and 2 above and the choice ofU that there is no y∈∂U such thatH(λ, y) =yfor λ∈[0,1].This shows thatH(λ,·) :U →C01(I) is an admissible homotopy; i.e. a compact homotopy without fixed points on∂U, the boundary ofU.

Therefore, H(λ,·) :U → C01(I) is an admissible homotopy between the constant map H(0,·) = 0 and the compact map H(1,·). Since 0 ∈ U, we have that H(0,·) is essential. By the topological transversality theorem of Granas, H(1,·) is essential.

This implies that it has a fixed point inU,and this fixed point is a solution of (1.21).

Since solutions of (1.21) are solutions of (1.1) we conclude that (1.1) has at least one solution, which is necessarily positive because of (1.4).

This completes the proof of the main result.

Remark. It is possible to obtain a uniqueness result if we assume, in addition to (H1) and (H2), the following condition:

(H3)There exists a constant M, such thatq(t) +M ≤q0π2,with strict inequality on a subset ofI with positive measure, andf(t, y1, z)−f(t, y2, z)≥ −M(y1−y2) for allt ∈I, z∈R and α ≤y2 ≤y1 ≤β.

Theorem 4.2 Assume that the conditions (H1), (H2) and (H3) hold. Then (1.1) has a unique positive solution y∈[α, β].

Proof. Theorem 4.1 guarantees the existence of at least one solutiony ∈[α, β].For contradiction, suppose there are two solutionsu, v ∈[α, β].

Assume first that u(t)≤v(t).Then

w(t) := v(t)−u(t)≥0 for all t∈I.

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Since

Lv(t)−Lu(t) =f(t, v(t), v0(t))−f(t, u(t), u0(t)) for allt∈I, assumption (H3) implies that

Lw(t)≥ −M w(t) for all t∈I, or

(L+M)w(t)≥0 for all t ∈I. (∗.1)

Suppose, next that v(t) ≤u(t) for all t ∈ I. Then −w(t) ≥ 0 for all t ∈I. This yields

(L+M)(−w(t))≥0 for all t∈I. (∗.2) Comparing (∗.1) and (∗.2) we see that (L+M)w(t) = 0 for all t∈I.The assump- tion onM and Lemma 3.3 imply that w(t) = 0 for all t∈I. Therefore

u(t) = v(t) for all t∈I.

This proves uniqueness.

5 Multiplicity of Solutions

In this section we use the previous result to get multiplicity of solutions of problem (1.1).

Theorem 5.1 Assume f : [0,1]×R2 → R is an L1-Caratheodory function and satisfies:

(H4) there are sequences {αj},{βj} of positive functions in C01(I) such that for all j = 1,2, . . . ,

(i) 0< αj < βj ≤αj+1,

(ii) Lαj(t)≥f(t, αj(t), α0j(t)), Lβj(t)≤ f(t, βj(t), βj0(t)), (iii) f(t, αj(t), α0j(t))>0> f(t, βj(t), βj0 (t)) , t ∈[0,1], (iv) the condition (H2) holds on [0,1]×[ αj, βj]×R.

Then (1.1) has infinitely many positive solutions yj such that αj ≤yj ≤βj.

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6 Example

Assume p(t) = 1 andq(t) = 0 for all t ∈[0,1] and consider the following problem, y00(t) =φ(t) (1 +y0(t)2)g(y(t)), 0< t <1,

y(0) =y(1) = 0, (1.9)

where φ∈L1(I) and g : R→Ris continuous and has an infinite number of positive simple zeros. This is the case if we assume the existence of an increasing sequence {aj}j∈N of positive numbers such that

g(aj) g(aj+1)<0 for j = 0,1, . . . .

We know of no applicable previous published works. However, f satisfies condition (H4) of Theorem 5.1, hence Problem (1.9) has infinitely many positive solutions.

Remark. A typical example for g is g(y) = siny, whose positive zeros form an infinite sequence {nπ; n= 1.2, ...}.

It is clear that the differential equation

y00(t) =φ(t) (1 +y0(t)2) sin(y(t)) has infinitely many positive solutions, yn(t) =nπ, n≥1.

The function f, defined by

f(t, y, z) = 2t(1 +z2) siny(t), 0≤t ≤1, changes sign infinitely many times. In fact we have

f(t, αj, z)>0 for αj = 1

2+ 2j

π, , j = 0,1,2, . . . , and

f(t, βj, z)<0 for βj = 3

2 + 2j

π, , j = 0,1,2, . . . .

Acknowledgement. The authors would like to thank an anonymous referee for comments that led to significant improvement of the original manuscript. Also, A. Boucherif is grateful to King Fahd University of Petroleum and Minerals for its constant support.

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(Received December 5, 2005)

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